Farm Tour

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14219		Accepted: 5421

Description

When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000. 

To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again. 

He wants his tour to be as short as possible, however he doesn't want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.

Input

* Line 1: Two space-separated integers: N and M. 

* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path's length. 

Output

A single line containing the length of the shortest tour. 

Sample Input

4 51 2 12 3 13 4 11 3 22 4 2

Sample Output

6

Source

 

农场主有N块地, 求从房子出发到某地,  再回到房子， 所用最短路程(每条边走一次)。 建源点和汇点 跑最短路。

数组开小为毛判T；

#include 
      
       #include 
       
        #include 
        
         using namespace std;typedef int Type;const int MAXNODE = 1010;const int MAXEDGE = 10010;const int INF = 0x3f3f3f3f;struct Edge{    int u, v, next;    Type cap, flow, cost;    Edge() {}    Edge(int u, int v, Type cap, Type flow, Type cost, int next):  u(u), v(v), cap(cap), flow(flow), cost(cost), next(next) {}};struct MXMF{    int n, m, s, t;    Edge edges[MAXEDGE*2];    int head[MAXNODE];    int p[MAXNODE];    Type dis[MAXNODE];    Type a[MAXNODE];    bool vis[MAXNODE];    void init(int n)    {        this->n=n;        memset(head, -1, sizeof(head));        m = 0;    }    void addEdge(int u, int v, Type cap, Type cost)    {        edges[m]=Edge(u, v, cap, 0, cost, head[u]);        head[u]=m++;        edges[m]=Edge(v, u, cap, 0, cost, head[v]);        head[v]=m++;    }    bool BellmanFord(int s, int t, Type &flow, Type &cost)    {        for(int i=0; i
         
           n+1;        memset(vis, 0, sizeof(vis));        dis[s]=0; vis[s]=true; p[s]=0; a[s]=INF;        queue
          
            Q;        Q.push(s);                 while(!Q.empty())        {            int u=Q.front(); Q.pop();            vis[u]=false;            for(int i=head[u]; i!=-1; i=edges[i].next)            {                Edge &e=edges[i];                if(e.cap>e.flow && dis[e.v]>dis[u]+e.cost)                {                    dis[e.v]=dis[u]+e.cost;                    p[e.v]=i;                    a[e.v]=min(a[u], e.cap-e.flow);                    if(!vis[e.v]) { Q.push(e.v); vis[e.v]=true;}                }            }        }        if(dis[t]== INF) return false;        flow += a[t];        cost += a[t]*dis[t];        int u=t;        while(u != s)        {            edges[p[u]].flow += a[t];            edges[p[u] ^ 1].cost = -edges[p[u] ^ 1].cost;            u = edges[p[u]].u;        }        return true;    }    Type minCost(int s, int t)    {        this->s=s; this->t;        Type flow=0, cost=0;        while(BellmanFord(s, t, flow, cost));        return cost;    }} mcmf; int n, m;void init(){    int source=1, sink=n;   //imp;     mcmf.init(n+2);    int u, v, c;    for(int i=0; i< m; i++)    {        scanf("%d%d%d", &u, &v, &c);        mcmf.addEdge(u, v, 1, c);    }    mcmf.addEdge(0, source, 2, 0);    mcmf.addEdge(sink, n+1, 2, 0);    printf("%d\n", mcmf.minCost(0, n+1));}int main(){    while(scanf("%d%d", &n, &m) != EOF){        init();    }    return 0;}